12/20/2015 Math 2 Honors - Santowski 1 Lesson 53 – Solving Trigonometric Equations Math 2 Honors -...

04/21/23 Math 2 Honors - Santowski 1

Lesson 53 – Solving Trigonometric Equations

Math 2 Honors - Santowski

(A) Review - Solving Equations The general strategy to ALGEBRAICALLY

solving ANY equations is to:

(a) Isolate the “base” function containing the unknown

(b) Isolate the unknown by using the inverse of the base function


(A) Review - Solving Equations And a second alternative to solving

ALGEBRAICALLY is to solve GRAPHICALLY by either:

(a) looking for an intersection point for f(x) = g(x)

(b) looking for the zeroes/roots of a rearranged eqn in the form of f(x) – g(x) = 0


(A) Examples

Here are some easy examples to start with Let’s use the domain -2π < x < 2π and then

work to an infinite domain


2)4csc(2 (f) 2

1)2sin( (e)

03)sin(2 (d) 01)tan( (c)

02)sec(3 (b) 01)sin( (a)

xx

xx

xx

(A) Examples - Graphically


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(B) Examples – With Calculator Solve the equation 3sin(x) – 2 = 0

We can rearrange as sin(x) = 2/3 so x = sin-1(2/3) giving us 41.8° (and the second angle being 180° - 41.8° = 138.2°)

Note that the ratio 2/3 is not one of our standard ratios corresponding to our “standard” angles (30,45,60), so we would use a calculator to actually find the related acute angle of 41.8°

04/21/23 Math 2 Honors - Santowski

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(B) Examples – With Calculator We can now solve the equation 3sin(x) – 2 = 0 by graphing f(x) = 3sin(x) – 2

and looking for the x-intercepts


(B) Examples – With Calculator Notice that there are 2 solutions

within the limited domain of 0° < < 360°

However, if we expand our domain, then we get two new solutions for every additional period we add

The new solutions are related to the original solutions, as they represent the positive and negative co-terminal angles

We can determine their values by simply adding or subtracting multiples of 360° (the period of the given function)


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(B) Examples – With Calculator Solve 4tan(2x) + 3 = 2 Again, we can set it up algebraically as tan(2x) = -1/4 and thus (2x)

= tan-1(-1/4) so x = ½ tan-1(-1/4) So thus x = ½ of 166° = 84° and x = ½ of 346° = 173°

To set it up graphically, we will make one minor change: we have two graphing options we can graph f(x) = 4tan(2x) + 1 and find the x-intercepts OR we can graph f(x) = 4tan(2x) + 3 and find out where f(x) = 2 to do this, we will simply graph y = 2 as a second equation and find out where the two graphs intersect


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(B) Examples – With Calculator


(B) Examples – With Calculator Notice that there are 2 solutions within the

limited domain of 0° < < 180° (NOTE: the period of a regular tan function is 180°)

However, if we expand our domain, then we get two new solutions for every additional period we add

The new solutions are related to the original solutions, as they represent the positive and negative co-terminal angles

We can determine their values by simply adding or subtracting multiples of 90° (the period of the given function the effect of the 2x in tan(2x) is a horizontal compression by a factor of 2, so we have reduced the period by a factor of 2)


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(C) Quadratic Trigonometric Equations Quadratic trig eqns contain terms like sin2(), cos2() Recall the Pythagorean identity (sin2() + cos2() = 1) Recall how to factor simple trinomials like x2 + 2x – 35 =

0 (x + 7)(x – 5) = 0 Recall how to factor difference of square trinomials like

4x2 – 25 = 0 (2x – 5)(2x + 5) = 0 Recall how to factor trinomials in the form of 3x2 – x – 4 =

0 using decomposition or guess & check (3x - 4)(x + 1) = 0


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(C) Quadratic Trigonometric Equations Solve 2cos2() = 1 if 0° < < 360 °


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(C) Quadratic Trigonometric Equations Solve 2cos2() = 1 if 0° < < 360 °

2 1

121

2

12

4 5 1 3 5 2 2 5 3 1 5

2

2

1

co s ( )

co s ( )

co s( )

co s

, , ,


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(C) Quadratic Trigonometric Equations Solve cos2(x) + 2cos(x) = 0 for 0 < x < 2


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(C) Quadratic Trigonometric Equations Solve cos2(x) + 2cos(x) = 0 for 0 < x < 2

co s ( ) co s( )

co s( ) co s( )

( ) co s( )

co s ( )

,

( ) co s( )

co s ( )

2

1

1

2 0

2 0

0

0

2

3

22

2

x x

x x

i x

x

x

ii x

x

x R


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(C) Quadratic Trigonometric Equations Solve 2cos2(x) - 3cos(x) + 1 = 0 for 0 < x < 2


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(C) Quadratic Trigonometric Equations Solve 2cos2(x) - 3cos(x) + 1 = 0 for 0 < x < 2

2 3 1 0

2 1 1 0

2 1 0

12

35

31

0 2

2co s ( ) co s( )

( co s( ) )(co s( ) )

( ) co s( )

co s( )

,

( ) co s( )

,

x x

x x

i x

x

x

ii x

x


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(C) Quadratic Trigonometric Equations Solve 2cos2(x) - sin(x) - 1 = 0


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(C) Quadratic Trigonometric Equations Solve 2cos2(x) - sin(x) - 1 = 0 notice we have both

sin(x) and cos(x) in the equation so use the Pythagorean identity to make changes

The equation becomes 2(1 – sin2(x)) – sin(x) – 1 = 0 So this is now –2sin2(x) – sin(x) + 1=0 or we can make it

0 = 2sin2(x) + sin(x) – 1 which we can now factor and solve on 0° < x < 360 °

(2sin(x) - 1)(sin(x) + 1) = 0 So 2sin(x) - 1 = 0 °, meaning x = 30 ° and 150 ° And sin(x) + 1 = 0 °, meaning x = 270 °


(D) Other Examples

Solve for x:


1)(sec2 )(

3)(sin4 )(

2

3

63cos )(

)sin()2sin( )(

0)cos(3)sin( )(

2

2

xe

xd

xc

xxb

xxa

Homework

HW

S14.6, p926, Q9,13,17,19,21,23,31,35,37,39,41,45,49,50


12/20/2015 Math 2 Honors - Santowski 1 Lesson 53 – Solving Trigonometric Equations Math 2 Honors -...

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